I have solved above question but I am not confident. I need guidnace if there is a mistake and how do I fix it.
To find an expression for $X_t$ using Itô's formula, we need to express $X_t$ as a function of a stochastic process. Let's define a function $F(t, W_t)$ such that $X_t$ can be derived from it using Itô's formula.
Given: $X_t = \int_0^t \sin(1 + 3W_s) dW_s$.
We define $F(t, W_t)$ such that $dF(t, W_t)$ corresponds to the integrand of $X_t$. A suitable choice for $F$ is a function whose derivative with respect to $W_t$ is $\sin(1 + 3W_t)$.
Let's consider $F(t, W_t) = -\frac{1}{3} \cos(1 + 3W_t)$. Then, applying Itô's formula:
$$ dF(t, W_t) = \frac{\partial F}{\partial t} dt + \frac{\partial F}{\partial W_t} dW_t + \frac{1}{2} \frac{\partial^2 F}{\partial W_t^2} (dW_t)^2. $$
Calculating the derivatives:
- $\frac{\partial F}{\partial W_t} = \sin(1 + 3W_t)$,
- $\frac{\partial^2 F}{\partial W_t^2} = 3 \cos(1 + 3W_t)$,
- $\frac{\partial F}{\partial t} = 0$ (since $F$ does not explicitly depend on $t$).
Substituting these into Itô's formula:
$$ dF(t, W_t) = \sin(1 + 3W_t) dW_t + \frac{1}{2} \cdot 3 \cos(1 + 3W_t) dt. $$
Integrating both sides from 0 to $t$:
$$ F(t, W_t) - F(0, W_0) = \int_0^t \sin(1 + 3W_s) dW_s + \frac{3}{2} \int_0^t \cos(1 + 3W_s) ds. $$
Since $F(0, W_0) = -\frac{1}{3} \cos(1)$, we have:
$$ X_t = F(t, W_t) + \frac{1}{3} \cos(1) - \frac{3}{2} \int_0^t \cos(1 + 3W_s) ds. $$
This is the expression for $X_t$ using Itô's formula.
For the mean and variance:
Mean of $X_t$:
- $\mathbb{E}[X_t] = \mathbb{E}[F(t, W_t)] + \frac{1}{3} \cos(1) - \frac{3}{2} \mathbb{E}\left[\int_0^t \cos(1 + 3W_s) ds\right]$.
- Since the expectation of an Itô integral is 0, $\mathbb{E}[X_t] = \frac{1}{3} \cos(1)$.
Variance of $X_t$:
- Using Itô isometry, $\text{Var}(X_t) = \mathbb{E}\left[\int_0^t \sin^2(1 + 3W_s) ds\right]$.